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Gravitational potential energy (roller coaster)

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data
    A 1000kg roller coaster, with its passengers, starts from rest at point A( h= 9.5 m) on a friction track whose profile is shown in the diagram above at right.
    a) what is its maximum speed?
    b) with what speed does the roller coaster arrive at point e?
    c) What constant braking force would have to be applied to the roller coaster at point E, to bring it to rest in a horizontal distance of 5.0 m
    Given data (from diagram)
    Point A- 9.5m
    Point B- 6.5m
    Point C- 9.2m
    Point D- 0.50m
    Point E- 5.5m

    2. Relevant equations
    Ep= mgh



    3. The attempt at a solution
    I used the gravitational potential energy equation to find gravitational potential energy at point A. I don't know if I need to get it for all the points! Also I don't understand where to start in this equation. I also wrote down the givens.
     

    Attached Files:

    Last edited: Dec 11, 2011
  2. jcsd
  3. Dec 11, 2011 #2

    NascentOxygen

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    Staff: Mentor

    Did you attach the diagram? Oh, I see, you provided the relevant heights.

    So it will be going fastest at the lowest point, assuming that is D. Is there another equation you know, relating energy to a moving body?
     
  4. Dec 11, 2011 #3
    For part A, Ep = mgh.
    Conservation of energy law states that the Ep at the top is equal to the Ek at the bottom. Ek = .5mv^2
    So, mgh = .5mv^2, where v is the velocity at the bottom.

    For the next part, I would use the conservation of energy law again. Except this time, you have velocity at the point where potential energy is the highest, so you have to account for the total mechanical energy of the system. At point B, which i'm assuming is at the top of another hill at 6.5m, Ep + Ek = Ep + Ek. The right side of the equation is for point B, and the left side of the equation is for the bottom of the hill after point A. mgh + .5mv^2 = mgh + .5mv^2., solve for v, and do so until you get to point e.

    For C, F=ma
    you know the mass, so now you need acceleration
    V = SQRT(2AD)
    solve for A, since D is 5 and V is the velocity you're going at point e that you just found

    then, F = ma
     
  5. Dec 11, 2011 #4
    Kinetic energy, work. I also know that Kinetic energy is equal to work.?
     
  6. Dec 11, 2011 #5
    The height in part a are the same right?
     
  7. Dec 11, 2011 #6
    V = SQRT(2AD). What does sqrt stand for? Sorry I have never used that!
     
  8. Dec 11, 2011 #7
    a) is it 99.05 m/s (also do i use the same height)
    b) for this I have to do every height there is?
    c) how is D 5? in the diagram you only have 5.5 and/or 0.5!
     
  9. Dec 11, 2011 #8

    NascentOxygen

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    Staff: Mentor

    How did you arrive at 99.05? Without explanation, who is to say whether you are right or not??

    In part c, the track levels out at a height of 5.5 metres. You want to bring the car to a halt in 5 metres along this level section, being the end of track.
     
    Last edited: Dec 11, 2011
  10. Dec 11, 2011 #9
    uh I found 99.05 by solving mgh= hv^2.
    (1000) (9.81) (0.5) = 0.5v^2
    4905/0.5= 0.5v^2/0.5
    Square(9810)= square(v^2)
    99.05m/s= v
     
  11. Dec 11, 2011 #10
    okay, now that I looked at the diagram lets approach this in another way.
    Let's find the Ep and Ek and velocities at each point, that way we can calculate the correct max speed, which will probably be at the lowest height, and get the right values for each point.

    so for part A, there is no velocity, so it's all potential.
    Ep = mgh = 93100 Joules

    Point B, the total mechanical energy of point A must equal the total mechanical energy of point B, B has Ep and Ek, because there is height and velocity

    mgh = .5mv^2 + mgh
    (1000)(9.8)(9.5) = (.5)(1000)(v^2) + (1000)(9.8)(6.5)
    93100 = 500v^2 + 63700
    29400 = (.5)(1000)v^2
    58.8 = v^2
    v = 7.668m/s at point B

    Moving onto point C, Energy at point B = .5mv^2 + mgh at point C
    (.5)(1000)(7.668^2) + (1000)(9.8)(6.5) = .5(1000)v^2 + (1000)(9.8)(9.2)
    v at point c = 15.065 m/s

    same thing for point D,
    .5mv^2 + mgh = .5mv^2 + mgh
    v = 19.936m/s

    and finally for point E,
    .5mv^2 + mgh = .5mv^2 + mgh
    v = 17.305m/s

    A) so, our max speed is at point D, which is 19.936
    B) we just got is, 17.305
    C) now we're moving in the horizontal. Our initial speed is 17.305 and our final speed is 0.
    F=MA
    to find acceleration, or in this case,V = SQRT(2AD)
    17.305 = the square root of (2*5*A)
    so, A = 29.947
    F = 29947N

    am I correct?
     
  12. Dec 11, 2011 #11
    sorry, F = MA , = 518.232N
     
  13. Dec 11, 2011 #12
    How did you get 518.232N? Like I know F= MA but what acceleration did you use?
     
  14. Dec 11, 2011 #13
    ... A = 29.947? the one we found on the last step
     
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